Angular dependence of electron-positron to two photons

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Hello,

First of all, it is not a homework question, just something I wonder about.

The dominating electron-positron annihilation process at low energies is photon pair creation. What is the angular dependence of the total cross section?

For some reason I expect the head-on collision to have the highest cross-section and the collinear situation to have the lowest cross-section, is it really the case?

I am not a particle physicist, but I expect the calculation to be pretty difficult, involving IR divergences and collinear divergences. Could you give me a reference to the solution of the problem or the experimental results and perhaps give some physical intuition on the whole thing.

Thanks in advance.
 

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  • #2
Meir Achuz
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"I expect the head-on collision to have the highest cross-section and the collinear situation to have the lowest cross-section"
I don't know what you mean by this.
In the center of mass system, it is always head-on.
In that system, the photon distribution is spherically symmetric.
The distribution in other initial configurations can be found by Lorentz transformation.
 
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Do I understand correctly that you claim that the invariant matrix element does not depend on the angle?
 
  • #4
Meir Achuz
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Yes. The electron and positron annihilate from a state of zero total spin and zero angular momentum so there is no angular distributon of the two photons.
 
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Vanadium 50
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But that's only half the answer. Why zero total spin?

That's more subtle - the answer there is that state has to produce three photons. So for exactly two photons, it's isotropic.
 

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